Characteristic subgroups, normal closures, and cores

Normality is not transitive: in $A_4$, $\langle (1\,2)(3\,4)\rangle \normal V_4$ and $V_4 \normal A_4$, but $\langle (1\,2)(3\,4)\rangle$ is not normal in $A_4$. A characteristic subgroup is invariant under every automorphism, and a characteristic subgroup of a normal subgroup of $G$ is normal in $G$.

Characteristic subgroups

Characteristic subgroups (Tag D-53LTN)

Let $G$ be a group. A subgroup $H\leq G$ is characteristic in $G$, written $H \ch G$, if $\phi(H) \leq H$ for every $\phi \in \Aut(G)$.

If $H \ch G$, then $\phi(H) = H$ for every $\phi\in\Aut(G)$: applying the definition to $\phi^{-1}$ gives $\phi^{-1}(H)\leq H$, so $H\leq\phi(H)$. The automorphisms need not fix $H$ pointwise.

(Characteristic and normal subgroups)

Every characteristic subgroup of $G$ is normal, because conjugation $\psi_g(x) = gxg^{-1}$ by $g \in G$ is an automorphism of $G$. A normal subgroup need not be characteristic: in the Klein four-group $V \cong C_2 \times C_2$, the subgroup $H = C_2 \times \{e\}$ is normal because $V$ is abelian, and the automorphism exchanging the two factors maps $H$ to $\{e\} \times C_2 \neq H$.

N/C theorem: $N_G(H)/C_G(H)$ embeds in $\Aut(H)$ (Tag PR-NESS4)

Normalizer modulo centralizer acts faithfully by conjugation on the subgroup.Suggest a slogan

Let $G$ be a group and $H\leq G$ a subgroup, with normalizer $N_G(H)$ and centralizer $C_G(H)$. The map $c\colon N_G(H)\to\Aut(H)$, $c(g)(h)=ghg^{-1}$, is a group homomorphism with kernel $C_G(H)$. Hence $C_G(H)\normal N_G(H)$ and $c$ induces an injective homomorphism \[ N_G(H)/C_G(H)\hookrightarrow\Aut(H). \]

For $g\in N_G(H)$ we have $gHg^{-1}=H$, so $c(g)$ is a bijection $H\to H$, and $c(g)(hh')=ghg^{-1}gh'g^{-1}=c(g)(h)\,c(g)(h')$; thus $c(g)\in\Aut(H)$. Since $c(gg')(h)=gg'h(gg')^{-1}=c(g)(c(g')(h))$, the map $c$ is a homomorphism. Its kernel is $\theset{g\in N_G(H)\suchthat ghg^{-1}=h\ \forall h\in H}=C_G(H)$. The first isomorphism theorem gives $N_G(H)/C_G(H)\cong c(N_G(H))\leq\Aut(H)$.

Characteristic subgroups of normal subgroups are normal (Tag PR-IOMVN)

Characteristic inside normal is normal in the ambient group.Suggest a slogan

Let $G$ be a group, $H \normal G$ a normal subgroup, and $N \ch H$ a characteristic subgroup of $H$. Then $N \normal G$.

Let $A \operatorname{char} B \normal C$, and let $c \in C$. Because $B \normal C$, conjugation $\psi_c(b) = cbc^{-1}$ restricts to an automorphism of $B$. Because $A$ is characteristic in $B$, $\psi_c(A) = A$. Hence $cAc^{-1} = A$ for every $c \in C$, and $A \normal C$.

The center is a characteristic subgroup (Tag PR-GJLQP)

Every automorphism preserves the center.Suggest a slogan

For every group $G$, the center $Z(G)$ is a characteristic subgroup of $G$, $Z(G) \ch G$.

Let $\phi \in \Aut(G)$, $z \in Z(G)$, and $g \in G$. Then $\phi(z)\,g = \phi\big(z\,\inverseof{\phi}(g)\big) = \phi\big(\inverseof{\phi}(g)\,z\big) = g\,\phi(z)$, so $\phi(z) \in Z(G)$.

Let $\psi \in \Aut(H)$ and $y\in Z(H)$. For every $h\in H$, \[ \begin{aligned} \psi(y)h &= \psi(y)\,\psi\big(\inverseof{\psi}(h)\big) \\ &= \psi\big( y \cdot \inverseof{\psi}(h) \big) \\ &= \psi\big( \inverseof{\psi}(h) \cdot y \big) && \text{since } \inverseof{\psi}(h)\in H \text{ and } y\in Z(H) \\ &= h\,\psi(y). \end{aligned} \] Hence $\psi(Z(H))\subseteq Z(H)$ for every $\psi\in\Aut(H)$.

Exercises

$Z(G)$ is characteristic in $G$ (Tag P-5UUNM)

Show that $Z(G) \leq G$ is always characteristic.

Solution

Let $\psi\in \Aut(G)$, $g\in Z(G)$, and $h\in G$. Since $g$ commutes with $\inverseof{\psi}(h)$, \[\psi(g)h=\psi(g)\,\psi(\inverseof{\psi}(h))=\psi(g\,\inverseof{\psi}(h))=\psi(\inverseof{\psi}(h)\,g)=h\,\psi(g).\] As $h$ is arbitrary, $\psi(g)\in Z(G)$, so $\psi(Z(G)) \subseteq Z(G)$. Applying the same argument to $\inverseof{\psi}$ yields $\inverseof{\psi}(Z(G)) \subseteq Z(G)$. Since $\psi$ is a bijection, $\psi\inverseof{\psi}(A) = A$ for all $A\leq G$, so $Z(G) \subseteq \psi(Z(G))$.

Normal closures and cores

Normal closure of a subgroup (Tag D-BPRD3)

Let $G$ be a group and $H\leq G$ a subgroup. The normal closure of $H$ in $G$ is the subgroup generated by the conjugates of $H$: \[ H^G \coloneqq \gens{ \Union_{g\in G} gHg^{-1} }. \]

Let $G$ be a group and $H\leq G$. Then $H^G$ is the smallest normal subgroup of $G$ containing $H$: \[ H^G = \bigcap \theset{N \st H \leq N \normal G }. \]

Normal core of a subgroup (Tag D-QMVEB)

Let $G$ be a group and $H\leq G$ a subgroup. The normal core of $H$ in $G$ is \[ H_G \coloneqq \Intersect_{g\in G} gHg^{-1}. \]

Let $H\leq G$, and let $\psi\colon G \to \Sym(G/H)$, $g \mapsto (xH\mapsto gxH)$, be the action of $G$ on the left cosets of $H$. Then $H_G$ is the largest normal subgroup of $G$ contained in $H$, and \[ H_G = \gens{ N \st N \normal G,\ N \leq H} = \ker \psi. \]

For $g\in G$, $\psi(g)=\id$ if and only if $gxH=xH$ for all $x\in G$, that is, $x^{-1}gx\in H$ for all $x$, that is, $g\in\Intersect_{x\in G}xHx^{-1}=H_G$. Hence $H_G=\ker\psi$ is a normal subgroup of $G$, and $H_G\subseteq 1\cdot H\cdot 1=H$. If $N\normal G$ and $N\leq H$, then $N=gNg^{-1}\subseteq gHg^{-1}$ for every $g$, so $N\subseteq H_G$. Therefore the subgroup generated by all such $N$ lies in $H_G$, and it contains $H_G$, which is one of them.

For $H\le G$, the normal closure of $H$ is the subgroup generated by the conjugates $gH\inverseof{g}$ for $g\in G$, and the normal core of $H$ is $\Intersect_{g\in G} gH\inverseof{g}$. The normal core of $H$ is the kernel of the action of $G$ on the left cosets $G/H$ by left multiplication, so it is trivial if and only if this action is faithful.

Frattini's argument (Tag T-4XKGD)

A normal subgroup together with the normalizer of one of its Sylow subgroups generates the whole group.Suggest a slogan

Let $G$ be a finite group, $H\normal G$ a normal subgroup, $p$ a prime, and $P \in \Syl_p(H)$ a Sylow $p$-subgroup of $H$. Then $G=H\,N_G(P)$, and $[G: H]$ divides $\abs{N_G(P)}$.

Let $g\in G$. Since $H\normal G$, $gPg^{-1}\leq H$ is a Sylow $p$-subgroup of $H$, so by Sylow's second theorem in $H$ there is $h\in H$ with $hgPg^{-1}h^{-1}=P$. Then $hg\in N_G(P)$, so $g\in h^{-1}N_G(P)\subseteq H\,N_G(P)$. Hence $G=H\,N_G(P)$, and by the second isomorphism theorem $G/H\cong N_G(P)/(N_G(P)\cap H)$, so $[G:H]$ divides $\abs{N_G(P)}$.

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